3z+2(6-z)=-(z+2)+2

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Solution for 3z+2(6-z)=-(z+2)+2 equation: 



3z+2(6-z)=-(z+2)+2

We move all terms to the left:

3z+2(6-z)-(-(z+2)+2)=0

We add all the numbers together, and all the variables

3z+2(-1z+6)-(-(z+2)+2)=0

We multiply parentheses

3z-2z-(-(z+2)+2)+12=0



We calculate terms in parentheses: -(-(z+2)+2), so:

-(z+2)+2

We get rid of parentheses

-z-2+2

We add all the numbers together, and all the variables

-z

Back to the equation:

-(-z)



We add all the numbers together, and all the variables

3z-2z-(-1z)+12=0

We add all the numbers together, and all the variables

z-(-1z)+12=0

We get rid of parentheses

z+1z+12=0

We add all the numbers together, and all the variables

2z+12=0

We move all terms containing z to the left, all other terms to the right

2z=-12

z=-12/2

z=-6

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